Answer: $y = 3x$

Integrate to recover f(x): Since $f'(x) = 6x^2 - 4x + 1$, integrating gives $f(x) = 2x^3 - 2x^2 + x + C$. Using $f(0) = 2$, we find $C = 2$, so $f(x) = 2x^3 - 2x^2 + x + 2$. Find the slope and point at x = 1: The slope is $f'(1) = 6 - 4 + 1 = 3$. The function value is $f(1) = 2 - 2 + 1 + 2 = 3$. The tangent passes through $(1, 3)$ with slope $3$. Write the tangent line equation: Point-slope form: $y - 3 = 3(x - 1)$, giving $y = 3x$. Option A is correct. Why distractors fail: Option B ($y = 3x + 2$) incorrectly uses $f(0) = 2$ as the $y$-intercept. Option C ($y = 3x - 2$) uses an incorrect $y$-intercept. Option D ($y = x + 2$) substitutes $f'(0) = 1$ for the slope instead of $f'(1) = 3$.

Answer: $-6$

Determine the required values: The tangent line at $x = 2$ needs the point $(2, F(2))$ and slope $F'(2)$. Compute F(2) and F'(2): $F(2) = \int_0^2 f(t)\,dt = 4$. By FTC Part 1, $F'(x) = f(x)$, so $F'(2) = f(2) = 5$. Find the y-intercept: Tangent line: $y - 4 = 5(x - 2)$, i.e., $y = 5x - 6$. At $x = 0$: $y = -6$. Option C is correct. Why distractors fail: Option A ($-10$) results from using the point $(2, 0)$ instead of $(2, 4)$. Option B ($4$) confuses $F(2)$ with the $y$-intercept. Option D ($5$) confuses the slope $F'(2) = 5$ with the $y$-intercept.

Answer: $\frac{2\sin(x^2)}{x}$

Apply FTC Part 1 combined with the Chain Rule: If $h(x) = \int_1^{u(x)} f(t)\,dt$, then $h'(x) = f(u(x)) \cdot u'(x)$. Here $u(x) = x^2$, $u'(x) = 2x$, and $f(t) = \frac{\sin(t)}{t}$. Compute the result: $h'(x) = \frac{\sin(x^2)}{x^2} \cdot 2x = \frac{2\sin(x^2)}{x}$. Option B is correct. Why distractors fail: Option A omits the chain rule factor $2x$. Option C substitutes $x$ rather than $x^2$ into the integrand. Option D correctly includes $2x$ but drops the denominator $t$ from the integrand, as if integrating $\sin(t)$ alone.

Answer: $b = 3$

Compute A(b) explicitly: $A(b) = \int_0^b (3x - x^2)\,dx = \frac{3b^2}{2} - \frac{b^3}{3}$. Find critical points via FTC: $A'(b) = f(b) = 3b - b^2 = b(3 - b)$. This equals zero at $b = 0$ and $b = 3$. For $0 < b < 3$, $f(b) > 0$ since the parabola $3x - x^2$ is positive between its roots $0$ and $3$. Therefore $A$ is strictly increasing on $(0, 3)$. Apply the Candidates Test: $A(0) = 0$ and $A(3) = \frac{27}{2} - 9 = \frac{9}{2}$. Since $A$ is increasing on $[0, 3]$, the absolute maximum is at $b = 3$. Option C is correct. Why distractors fail: Option A ($b = 0$) gives the minimum $A(0) = 0$. Option B ($b = 3/2$) is where $f$ itself reaches its peak value, but this maximizes the integrand, not the accumulated area. Option D ($b = 2$) gives $A(2) = 6 - 8/3 = 10/3 \approx 3.33$, which is less than $A(3) = 9/2 = 4.5$.

Answer: $\frac{dy}{dx} = \frac{2x}{1+\cos(y)}$;\ area $= \int_0^a y\,dx$ (with $y$ implicitly defined);\ arc length $= \int_0^a \sqrt{1 + \left(\frac{2x}{1+\cos(y)}\right)^{2}}\,dx$

Implicit differentiation: Differentiating $y + \sin(y) = x^2$ with respect to $x$: $\frac{dy}{dx} + \cos(y)\frac{dy}{dx} = 2x$, so $\frac{dy}{dx}(1 + \cos(y)) = 2x$, giving $\frac{dy}{dx} = \frac{2x}{1+\cos(y)}$. Area under the curve: The area under a curve $y(x) \geq 0$ from $0$ to $a$ is $\int_0^a y\,dx$. Although $y$ is only implicitly defined, the integral expression is still $\int_0^a y\,dx$. Arc length setup: Arc length $= \int_0^a \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx = \int_0^a \sqrt{1 + \left(\frac{2x}{1+\cos(y)}\right)^2}\,dx$. Option A correctly combines all three results. Why distractors fail: Option B omits the $+1$ in $1 + \cos(y)$, producing an incorrect derivative. Option C replaces $y$ with $x^2$ in the area integral (ignoring $\sin(y)$) and oversimplifies the arc length by removing the implicit dependence on $y$. Option D incorrectly solves for $\frac{dy}{dx}$ by treating it algebraically rather than factoring.

Answer: The particle changes direction at $t = 1$ and $t = 3$, and the total distance traveled exceeds the net displacement.

Find when velocity is zero: $v(t) = t^2 - 4t + 3 = (t - 1)(t - 3) = 0$ at $t = 1$ and $t = 3$. The velocity changes sign at both roots. Analyze the sign of velocity: For $0 < t < 1$: $v(t) > 0$ (positive direction). For $1 < t < 3$: $v(t) < 0$ (negative direction). For $3 < t < 4$: $v(t) > 0$ (positive direction). The particle reverses direction twice. Compare distance and displacement: Because the velocity changes sign, $\int_0^4 |v(t)|\,dt > \left|\int_0^4 v(t)\,dt\right|$. Total distance exceeds net displacement. Option B is correct. Why distractors fail: Option A ignores sign changes. Option C confuses the vertex ($t = 2$) with the roots and falsely equates distance with displacement. Option D identifies the turning points correctly but incorrectly claims distance equals displacement.

Answer: $\frac{4}{17}$

Apply the Fundamental Theorem of Calculus Part 1: By FTC Part 1, if $g(x) = \int_0^x f(t)\,dt$, then $g'(x) = f(x)$. Here $f(t) = \frac{t}{t^2+1}$, so $g'(x) = \frac{x}{x^2+1}$. Evaluate at x = 4: $g'(4) = \frac{4}{16 + 1} = \frac{4}{17}$. Option B is correct. Why distractors fail: Option A ($\frac{1}{17}$) evaluates $\frac{1}{x^2+1}$ at $x = 4$, dropping the numerator $x$. Option D ($\frac{1}{2}\ln(17)$) computes the integral $g(4)$ via $u$-substitution instead of the derivative $g'(4)$. Option C ($\ln(17)$) doubles this integral value.

Answer: Both applications are valid; after the second, the limit simplifies to $\displaystyle\lim_{x \to 0}\frac{2xe^{x^2}}{6x} = \frac{1}{3}$.

Verify the initial indeterminate form: At $x = 0$: numerator $= \int_0^0(e^{t^2}-1)\,dt = 0$ and denominator $= 0^3 = 0$, so we have $\frac{0}{0}$. First application of L'Hôpital's Rule: By FTC Part 1, $\frac{d}{dx}\int_0^x (e^{t^2}-1)\,dt = e^{x^2}-1$. The derivative of $x^3$ is $3x^2$. The new limit $\lim_{x \to 0}\frac{e^{x^2}-1}{3x^2}$ is again $\frac{0}{0}$, so a second application is justified. Second application and final answer: Differentiating: numerator gives $2xe^{x^2}$, denominator gives $6x$. So $\lim_{x \to 0}\frac{2xe^{x^2}}{6x} = \lim_{x \to 0}\frac{2e^{x^2}}{6} = \frac{2}{6} = \frac{1}{3}$. Option C is correct. Why distractors fail: Option A falsely claims FTC and L'Hôpital's Rule cannot be combined. Option B incorrectly asserts $\frac{e^{x^2}-1}{3x^2}$ is not indeterminate at $x = 0$ (it is $\frac{0}{0}$). Option D gets $\frac{2}{3}$, likely from a differentiation error.